Chemistry Calculating Final Composition of Compounds in Haber's Process with 50% Yield

AI Thread Summary
The discussion revolves around calculating the final composition of compounds in the Haber process with a 50% yield from 30 liters of N2 and 30 liters of H2. Initially, the reaction is set up as N2 + 3H2 -> 2NH3, leading to a theoretical production of ammonia. However, with a 50% yield, it is clarified that only half of the maximum theoretical ammonia is produced, resulting in 10 liters of NH3. Consequently, the remaining amounts are 15 liters of H2 and 25 liters of N2, confirming that hydrogen is the limiting reagent. The conversation emphasizes the importance of stoichiometry in determining the final quantities of reactants and products.
Suyash Singh
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Homework Statement


30 litres of N2 and 30 litres of H2 are taken in Habers process with 50% yield.What is the final composition of the compunds in litres?

Homework Equations


N2+3H2->2NH3

The Attempt at a Solution


N2 + 3H2-> 2NH3
intial: 30 30 0
final: 30 - x/2 30-(3x)/2 (2x)/2

so,
30 - x/2 + 30-(3x)/2 = (2x)/2

this gives x=20
N2=20L
H2=0L
NH3=20L

Am i right?
 
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Suyash Singh said:
so,
30 - x/2 + 30-(3x)/2 = (2x)/2
Why do you expect these two to be equal?

How is the yield defined? Using all hydrogen with just 50% yield is odd.
 
mfb said:
Why do you expect these two to be equal?

Howis the yield defined? Using all hydrogen with just 50% yield is odd.
litres are conserved
50 percent yield means half of theoretical value of NH3
 
Suyash Singh said:
litres are conserved

2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?

Suyash Singh said:
50 percent yield means half of theoretical value of NH3

If only 50% of ammonia was produced, was all the limiting reagent consumed?
 
Borek said:
2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?
If only 50% of ammonia was produced, was all the limiting reagent consumed?
Ok i understand the limiting concept now.

NH3 produced=2/3 x 30 =20 L

H2 left= 0 L

N2 left= 30- (1/2 x 20) =20 L
 
So which one is the limiting reagent here? And how much was left if only 50% reacted?
 
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Borek said:
So which one is the limiting reagent here? And how much was left if only 50% reacted?
H2 the limiting reagent

all the elements react only 50 %?
 
Suyash Singh said:
H2 the limiting reagent

Right.

all the elements react only 50 %?

Stop guessing, use stoichiometry to calculate.
 
Suyash Singh said:
50 percent yield means half of theoretical value of NH3
But you got all of the maximum yield of NH3: You used up all hydrogen.
Based on that you can figure out how much NH3 is produced if you use only half of it.
 
  • #10
NH3 produced = 50/100 x (2/3 x 30) = 10 L
H2 left
= 30 - (3/2 x 10) = 15 L
N2 left
= 30 - (1/2 x 10)= 25 L
 
  • #11
That looks good.
 
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